41

I've been playing with beautiful soup and parsing web pages for a few days. I have been using a line of code which has been my saviour in all the scripts that I write. The line of code is : 

r = requests.get('some_url', auth=('my_username', 'my_password')).

BUT ...

I want to do the same thing with (OPEN A URL WITH AUTHENTICATION):

(1) sauce = urllib.request.urlopen(url).read() (1)
(2) soup = bs.BeautifulSoup(sauce,"html.parser") (2)

I'm not able to open a url and read, the webpage which needs authentication. How do I achieve something like this :

  (3) sauce = urllib.request.urlopen(url, auth=(username, password)).read() (3) 
instead of (1)

4 Answers4

32

You're using HTTP Basic Authentication:

import urllib2, base64

request = urllib2.Request(url)
base64string = base64.b64encode('%s:%s' % (username, password))
request.add_header("Authorization", "Basic %s" % base64string)   
result = urllib2.urlopen(request)

So you should base64 encode the username and password and send it as an Authorization header.

moritzg
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23

Have a look at the HOWTO Fetch Internet Resources Using The urllib Package from the official docs:

# create a password manager
password_mgr = urllib.request.HTTPPasswordMgrWithDefaultRealm()

# Add the username and password.
# If we knew the realm, we could use it instead of None.
top_level_url = "http://example.com/foo/"
password_mgr.add_password(None, top_level_url, username, password)

handler = urllib.request.HTTPBasicAuthHandler(password_mgr)

# create "opener" (OpenerDirector instance)
opener = urllib.request.build_opener(handler)

# use the opener to fetch a URL
opener.open(a_url)

# Install the opener.
# Now all calls to urllib.request.urlopen use our opener.
urllib.request.install_opener(opener)
Christian König
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  • Please see the line (2) in the question. I need to parse the sauce using beautiful soup. How do I achieve that using your code? –  May 29 '17 at 10:44
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    `AbstractBasicAuthHandler does not support the following scheme: 'Bearer'` – aggregate1166877 Aug 17 '19 at 23:55
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    @ChristianKönig `Python38\lib\urllib\request.py", line 1014, in http_error_auth_reqed raise ValueError("AbstractBasicAuthHandler does not " ValueError: AbstractBasicAuthHandler does not support the following scheme: 'Digest'` – Bilal Apr 23 '21 at 11:28
6

With urllib3 :

import urllib3

http = urllib3.PoolManager()
myHeaders = urllib3.util.make_headers(basic_auth='my_username:my_password')
http.request('GET', 'http://example.org', headers=myHeaders)
Skippy le Grand Gourou
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1

Use this. This is standard urllib found with Python3 installation. Works great guaranteed. Also, see gist

import urllib.request

url = 'http://192.168.0.1/'

auth_user="username"
auth_passwd="^&%$$%^"

passman = urllib.request.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, url, auth_user, auth_passwd)
authhandler = urllib.request.HTTPBasicAuthHandler(passman)
opener = urllib.request.build_opener(authhandler)
urllib.request.install_opener(opener)

res = urllib.request.urlopen(url)
res_body = res.read()
print(res_body.decode('utf-8'))
Apurva Singh
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